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Consider the set $\Omega = \{(x,y,z) \in \mathbb{R}^3: \sqrt{x^2+y^2}\leq z< 1\} $

The question said to show that it is neither closed nor open, but I am getting an apparent contradiction.

Define $$ \begin{aligned} f&: \mathbb{R}^2 &\hspace{-0.8cm}\longrightarrow &\,\mathbb{R}^3 \\ &(x,y) &\hspace{-0.65cm}\mapsto &\Big(x,y,\sqrt{x^2+y^2}\,\Big) \end{aligned} $$ which is continuous (I think?).

Then, the preimage $$f^{-1}(\Omega)=\{{(x,y)\in \mathbb{R}^{3}:\sqrt{x^2+y^2}< 1}\}$$

Now, I am pretty sure I can easily show that this pre image is an open set in $\mathbb{R}^2$ (correct me if I'm wrong about this), so isn't this a contradiction? Since there's a theorem which states that if a function $\mathbb{R}^m \to\mathbb{R}^n$ is continuous, then any the pre image of an open set must be open in $\mathbb{R}^m$ too. So this implies that if a pre image is not open, the image is not open (which is what I'm trying to use to show that the set isn't open), but there seems to be an apparent contradiction in my working out because clearly the pre image is an open set. Could someone help me out here?

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The preimage of $\Omega$ is open does not imply that $\Omega$ is open, you have to write $\Omega=g^{-1}(U)$ where $g$ is continuous and $U$ open.

$\Omega$ is not open since its intersection with a plane $\{z=c \}$ where $0<c<1$ is not open in that plane.

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  • $\begingroup$ I am not sure where I made the assumption that pre image is open => omega is open? The assumption I made was that if the pre image of a continuous is NOT open, then Omega cannot be open. Thank you for your explanation of why it's not open, but can you point out where I was wrong in my opening post? (Unless my first sentence here does not address your criticism) $\endgroup$ – user542938 Mar 21 '18 at 12:12
  • $\begingroup$ Oh wait now I see what you mean. Either way, I am looking for a method that is something along the lines of: choosing a suitable function f, and showing that f^-1 (Omega) is not open. $\endgroup$ – user542938 Mar 21 '18 at 12:20

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